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Quadratic Equations

Chapter 16. Quadratic Equations. Chapter Sections. 16.1 – Solving Quadratic Equations by the Square Root Property 16.2 – Solving Quadratic Equations by Completing the Square 16.3 – Solving Quadratic Equations by the Quadratic Formula 16.4 – Graphing Quadratic Equations in Two Variables

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Quadratic Equations

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  1. Chapter 16 Quadratic Equations

  2. Chapter Sections 16.1 – Solving Quadratic Equations by the Square Root Property 16.2 – Solving Quadratic Equations by Completing the Square 16.3 – Solving Quadratic Equations by the Quadratic Formula 16.4 – Graphing Quadratic Equations in Two Variables 16.5 – Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions

  3. Solving Quadratic Equations by the Square Root Property § 16.1

  4. Square Root Property We previously have used factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a2 = b, then

  5. Square Root Property Example Solve x2 = 49 Solve 2x2 = 4 x2 = 2 Solve (y – 3)2 = 4 y = 3  2 y = 1 or 5

  6. Square Root Property Example Solve x2 + 4 = 0 x2 = 4 There is no real solution because the square root of 4 is not a real number.

  7. Square Root Property Example Solve (x + 2)2 = 25 x = 2 ± 5 x = 2 + 5 or x = 2 – 5 x = 3 or x = 7

  8. Square Root Property 3x – 17 = Example Solve (3x – 17)2 = 28

  9. Solving Quadratic Equations by Completing the Square § 16.2

  10. Completing the Square In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).

  11. add Completing the Square Example What constant term should be added to the following expressions to create a perfect square trinomial? • x2 – 10x add 52 = 25 • x2 + 16x add 82 = 64 • x2 – 7x

  12. Completing the Square Example We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.

  13. Completing the Square Solving a Quadratic Equation by Completing a Square • If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. • Isolate all variable terms on one side of the equation. • Complete the square (half the coefficient of the x term squared, added to both sides of the equation). • Factor the resulting trinomial. • Use the square root property.

  14. y + 3 = ± = ± 1 Solving Equations Example Solve by completing the square. y2 + 6y = 8 y2 + 6y+ 9 = 8 + 9 (y + 3)2 = 1 y = 3 ± 1 y = 4 or 2

  15. (y + ½)2 = Solving Equations Example Solve by completing the square. y2 + y – 7 = 0 y2 + y = 7 y2 + y+ ¼ = 7 + ¼

  16. x2 + 7x + = ½ + = (x + )2 = Solving Equations Example Solve by completing the square. 2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½

  17. Solving Quadratic Equations by the Quadratic Formula § 16.3

  18. The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.

  19. The Quadratic Formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.

  20. The Quadratic Formula Example Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1

  21. Solve x2 + x – = 0 by the quadratic formula. The Quadratic Formula Example x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20

  22. The Quadratic Formula Example Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.

  23. The Discriminant The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

  24. The Discriminant Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions.

  25. Solving Quadratic Equations Steps in Solving Quadratic Equations • If the equation is in the form (ax+b)2 = c, use the square root property to solve. • If not solved in step 1, write the equation in standard form. • Try to solve by factoring. • If you haven’t solved it yet, use the quadratic formula.

  26. Solving Equations Example Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1) Let a = 1, b = -3, c = 1

  27. Solving Equations Example Solve the following quadratic equation.

  28. Graphing Quadratic Equations in Two Variables § 16.4

  29. Graphs of Quadratic Equations We spent a lot of time graphing linear equations in chapter 3. The graph of a quadratic equation is a parabola. The highest point or lowest point on the parabola is the vertex. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola.

  30. y xy x Graphs of Quadratic Equations Example Graph y = 2x2 – 4. (2, 4) (–2, 4) 2 4 1 –2 (–1, – 2) (1, –2) 0 –4 –1 –2 (0, –4) –2 4

  31. Intercepts of the Parabola Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y.

  32. 2) the x-coordinate of the vertex is . Characteristics of the Parabola If the quadratic equation is written in standard form, y = ax2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.

  33. y Since a = –2 and b = 4, the graph opens down and the x-coordinate of the vertex is xy x Graphs of Quadratic Equations Example Graph y = –2x2 + 4x + 5. (1, 7) (2, 5) (0, 5) 3 –1 (–1, –1) (3, –1) 5 2 1 7 0 5 –1 –1

  34. Interval Notation, Finding Domain and Ranges from Graphs, and Graphing Piecewise-Defined Functions § 16.5

  35. Domain and Range Recall that a set of ordered pairs is also called a relation. The domainis the set of x-coordinates of the ordered pairs. The rangeis the set of y-coordinates of the ordered pairs.

  36. Domain and Range Example Find the domain and range of the relation {(4,9), (–4,9), (2,3), (10, –5)} • Domain is the set of all x-values, {4, –4, 2, 10} • Range is the set of all y-values, {9, 3, –5}

  37. y Domain x Range Domain is [–3, 4] Range is [–4, 2] Domain and Range Example Find the domain and range of the function graphed to the right. Use interval notation.

  38. y Range x Domain is (– , ) Range is [– 2, ) Domain Domain and Range Example Find the domain and range of the function graphed to the right. Use interval notation.

  39. Input (Animal) Polar Bear Cow Chimpanzee Giraffe Gorilla Kangaroo Red Fox Output (Life Span) 20 15 10 7 Domain and Range Example Find the domain and range of the following relation.

  40. Domain and Range Example continued Domain is {Polar Bear, Cow, Chimpanzee, Giraffe, Gorilla, Kangaroo, Red Fox} Range is {20, 15, 10, 7}

  41. Graph Graphing Piecewise-Defined Functions Example Graph each “piece” separately. Values  0. Values > 0. Continued.

  42. y (3, 6) Open circle (0, 3) x (0, –1) (–1, 4) (–2, 7) Graphing Piecewise-Defined Functions Example continued

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